Using second law of motion, derive the relation between force and acceleration. A bullet of 10 g strikes a sand-bag at a speed of 103 m s-1 and gets embedded after travelling 5 cm. Calculate
(i) the resistive force exerted by the sand on the bullet
(ii) the time taken by the bullet to come to rest.
Mass of bullet = 10 g = 10/1000 = 0.01 kg
Initial velocity, u = 10^3 m/s
Final velocity, v = 0 m/s
Distance, s = 5 cm = 5/100 m
Using second law of motion,
(i) v^2 - u^2 = 2 as
0^2 - (10^3)^2 = 2.a.(5/100)
a = (-1000 x 1000 x 100) / 2 x 5 = - 10^7 m/s^2
F = ma = 10^5 N
(ii) v = u + at
0 = 10^3 - 10^7 t
10^7 t = 10^3, t = 10^3 / 10^7 = 1/10^4 sec.
Time taken by the bullet to come to rest is 1/10^4 second.
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Welcome to the NCERT Solutions for Class 9 Science - Chapter . This page offers a step-by-step solution to the specific question from Excercise 0 , Question 18: Using second law of motion, derive the relation between force and acceleration. A bullet of 10 g str....
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